Let the circle be x^2+y^2+2gx+2fy+c=0,
let us seek normal and tangent at (x_1,y_1)
(note that x_1^2+y_1^2+2gx_1+2fy_1+c=0) ...(A)
as the center of the circle is (-g,-f),
as normal joins (-g,-f) and (x_1,y_1) its slope is (y_1+f)/(x_1+g)
and equation of normal is y-y_1=(y_1+f)/(x_1+g)(x-x_1)
i.e. (x_1+g)y-x_1y_1-gy_1=(y_1+f)x-x_1y_1-fx_1
or (y_1+f)x-(x_1+g)y-fx_1+gy_1=0
or (y_1+f)x-(x_1+g)y-fx_1+gy_1=0
and as product of slopes of normal and tangent is -1,
slope of tangent is -(x_1+g)/(y_1+f) and its equation will be
y-y_1=-(x_1+g)/(y_1+f)(x-x_1)
or (y-y_1)(y_1+f)+(x-x_1)(x_1+g)=0
or yy_1-y_1^2+fy-fy_1+x x_1-x_1^2+gx-gx_1=0
or x x_1+yy_1+gx+fy-(x_1^2+y_1^2+fy_1+gx_1)=0
from (A) x_1^2+y_1^2+fy_1+gx_1=-fy_1-gx_1-c
Hence equation of tangent is x x_1+yy_1+gx+fy+fy_1+gx_1+c=0
or x x_1+yy_1+g(x+x_1)+f(y+y_1)+c=0
